Problems

Solution:

1. \begin{align*} && I &= \int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx \\ x = u^{-1}, \d x = -u^{-2} \d u: && &= \int_{u=\infty}^{u = 0} \frac{u^{-1/2} - 1}{\sqrt{u^{-1}(u^{-3}+1)}} (-u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-1/2} -1}{\sqrt{1+u^3}} \d u \\ &&&= \int_0^\infty \frac{1-\sqrt{u}}{\sqrt{u(u^3+1)}} \d u \\ &&&= -I \\ \Rightarrow && I &= 0 \end{align*}
2. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx \\ x = u^{-2}, \d x = -2u^{-3} \d u: && &= \int_{u = \infty}^{u = 0} \frac{1}{\sqrt{u^{-6}+1}} (-2u^{-3}) \d u \\ &&&= \int_0^\infty \frac{2}{\sqrt{1+u^6}} \d u \\ &&&= 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, \d x \end{align*}
3. \begin{align*} && I &= \int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx \\ x = u^{-1}, \d x = - u^{-2} \d t: &&&= \int_{u=\infty}^{u = 0} \frac{u^{-r}-1}{\sqrt{u^{-s}(u^{-p}+1)}} (- u^{-2}) \d u \\ &&&= \int_0^\infty \frac{u^{-r}-1}{\sqrt{u^{4-s}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r}(u^{-p}+1)}} \d u \\ &&&= \int_0^\infty \frac{1-u^{r}}{\sqrt{u^{4-s+2r-p}(1+u^p)}} \d u \\ \end{align*} Therefore if \(4-s+2r-p = s\) or \(r = \frac{p}2+s-2\) we have \(I = -I\), ie \(I = 0\).
4. \begin{align*} && I &= \int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx \\ x = u^{-t}, \d x = -t u^{-t-1}:&&&= \int_{u = -\infty}^{u = 0} \frac{1}{\sqrt{u^{-pt}+1}} (-t u^{-t-1}) \d u \\ &&&= t \int_0^\infty \frac1{\sqrt{u^{-pt+2t+2}+u^{2t+2}}} \d u \end{align*} Therefore if \(\begin{cases} 2t+2 &= q \\ 2t+2 -pt &= 0 \end{cases}\), ie \(q = 2t+2, p = 2 + 2/t\) we will have found the \(p, q\) desired for any \(t\) (or \(k\)). [Alternatively] Let \(\displaystyle I(p) =\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx\), then clearly \(I(p)\) is decreasing and as \(p \to \infty\) \(I(p) \to 1\), so our integral can take any values on \((1, \infty)\) and so for any positive value we can find two values with a given ratio. In particular given \(k\) and \(p\) we can find a suitable \(q\).

Solution:

1. Suppose the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers then \begin{align*} && \sum_{i=c}^{i=c+n+k-1} i &= \sum_{i=c+n+k}^{c+2n+k-1} i \\ \Leftrightarrow && \frac{(c+n+k-1)(c+n+k)}{2} - \frac{(c-1)c}{2} &= \frac{(c+2n+k-1)(c+2n+k)}{2} - \frac{(c+n+k-1)(c+n+k)}{2} \\ \Leftrightarrow && 2(c+n+k-1)(c+n+k) &= (c+2n+k-1)(c+2n+k) + c(c-1) \\ \Leftrightarrow && 2c^2+4cn+4ck+2n^2+4kn+2k^2-2c-2n-2k&=2c^2+4cn+2ck+4n^2+4nk+k^2-2c-2n-k \\ \Leftrightarrow && 2ck+k^2&=2n^2+k \\ \end{align*}
2. 1. If \(k=1\) then \begin{align*} && 2n^2 + 1 &= 2c + 1 \\ \Rightarrow && c &= n^2 \end{align*} So \(n\) can take any value and \(c = n^2\)
   2. If \(k=2\) then \begin{align*} && 2n^2+2&= 4c+4 \\ \Rightarrow && n^2-1 &=2c \end{align*} So \(n\) must be odd, and \(c = \frac12(n^2-1)\)
3. Suppose \(k=4\) then \(2n^2+4 = 8c+16\) or \(n^2-6 = 4c\) but then the left hand side is \(2, 3 \pmod{4}\) which is a contradiction.
4. Suppose \(c =1\)
   1. Since \(2n^2+k = 2k + k^2\) or \(2n^2 = k^2+k\) we can have \(k = 1, n = 1\) or \(k = 8, n = 6\)
   2. Suppose \(2N^2 = K^2 + K\) then consider \begin{align*} && 2(N')^2 &= 2(3N+2K+1)^2 \\ &&&= 2(9N^2+4K^2+1+12NK+6N+4K) \\ &&&= 18N^2+8K^2+24NK+12N+8K+2 \\ && (K')^2+K' &= (4N+3K+1)^2 + (4N+3K+1) \\ &&&= 16N^2 + 9K^2+1+24NK+12N+9K+1 \\ &&&= 16N^2+9K^2+24NK+12N+9K+2 \\ \Rightarrow && 2(N')^2-(K')^2-K' &= 2N^2-K^2-K \\ &&&= 0 \end{align*} as required.
   3. So consider \((k,n) = (1,1), (8,6), (49, 35), (288,204)\)

Solution:

1. \(\,\) \begin{align*} && 3 &= (x-\sqrt2)^2 \\ &&&= x^2 - 2\sqrt2 x + 2 \\ \Rightarrow && 2\sqrt2 x &= x^2-1 \\ \Rightarrow && 8x^2 &= x^4 - 2x^2 + 1 \\ \Rightarrow && 0 &= x^4 - 10x^2 + 1 \end{align*} Noticing that \((\sqrt2+\sqrt3-\sqrt2)^2 = 3\) we note that \(\sqrt2 + \sqrt3\) is a root of our quartic.
2. Suppose \(x = \sqrt2 + \sqrt3 + \sqrt5\) then \begin{align*} && 0 &= (x - \sqrt5)^4 - 10(x-\sqrt5)^2 + 1 \\ &&&= x^4 - 4\sqrt5x^3 + 30x^2-20\sqrt5 x +25 - 10x^2+20\sqrt5x -50 + 1\\ &&&= (x^4+20x^2- 24) - 4\sqrt5 x^3 \\ \Rightarrow && 80x^6 &= (x^4+20x^2-24)^2 \\ &&&= x^8 + 40x^6 + 352x^4 - 960x^2+576 \\ \Rightarrow && 0 &= x^8-40x^6 + 352x^4-960x^2+576 \end{align*} So take \(g(x) = x^8-40x^6 + 352x^4-960x^2+576\).
3. Notice that if \(p(t) = t^3-3t+1\) then \(p(t -\sqrt2) = 0\) for \(t = a,b,c\) so \begin{align*} && 0 &= (t - \sqrt2)^3 -3(t - \sqrt2) + 1 \\ &&&= t^3-3\sqrt2 t^2 + 6t - 2\sqrt2 - 3t + 3\sqrt 2 + 1 \\ &&&= (t^3+3t+1) - \sqrt2 (3t^2+1) \\ \Rightarrow && 2(3t^2+1)^2 &= (t^3+3t+1)^2 \\ \Rightarrow && 2(9t^4+6t^2+1) &= t^6 + 6t^4+2t^3+9t^2+6t+1 \\ \Rightarrow && 0 &= t^6-12t^4+2t^3-3t^2+6t-1 \end{align*}
4. \(\,\) \begin{align*} && t &= \sqrt[3]{2} + \sqrt[3]{3} \\ \Rightarrow && t^3 &= 2 + 3\sqrt[3]{12} + 3\sqrt[3]{18} + 3 \\ &&&= 5 + 3 \sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3}) \\ &&&= 5 + 3\sqrt[3]{6}t \\ \Rightarrow && 162t^3 &= (t^3-5)^3 \\ &&&= t^9-15t^6+75t^3 -125 \\ \Rightarrow && 0 &= t^9-15t^6-87t^3-125 \end{align*} so \(k(x) = x^9 - 15x^6-87x^3-125\)

Solution:

1. \(\,\) \begin{align*} && \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\ &&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align*} Notice since \(a,b\) and \(c\) are distinct real numbers the vector \(\langle a,b,c \rangle\) cannot be the zero vector, so the determinant of the matrix is zero, ie \(0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx\). Notice also then that \begin{align*} && \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\ &&&= (x+y+z)^2 - 2(xy+yz+zx) \\ &&&= 2 + (x+y+z)^2 \geq 2 \end{align*}
2. \(\,\) \begin{align*} && \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} 2a-xb-xc \\ -ay+2b-yc \\ -za-zb+2c \end{pmatrix} \\ &&&= \begin{pmatrix} 2a-x(b+c) \\ 2b-y(c+a) \\ 2c-z(a+b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \Rightarrow && 0 &= \det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \\ &&&= 2(4 -yz)+x(-2y-yz)-x(yz+2z) \\ &&&= 8 - 2yz-2yx-2xyz-2zx\\ \Rightarrow && 4 &= xyz+xy+yz+zx \end{align*} \begin{align*} && (2a + b + c)(a + 2b + c)(a + b + 2c) &> 5(b+c)(c+a)(a+b) \\ \Leftrightarrow && \left ( \frac{2a}{b+c}+1 \right)\left ( \frac{2b}{c+a}+1 \right)\left ( \frac{2c}{a+b}+1 \right) &> 5 \\ \Leftrightarrow && \left ( x+1 \right)\left ( y+1 \right)\left ( x+1 \right) &> 5 \\ \Leftrightarrow && xyz+xy+yz+zx+x+y+z+1 &> 5 \\ \Leftrightarrow && 5+x+y+z&> 5 \\ \end{align*} Which is clearly true since if \(a,b,c\) are positve real numbers so are \(x,y,z\). This final inequality is equivalent to showing \(x+y+z > 2\) ie \begin{align*} && x+y+z &> 2 \\ \Leftrightarrow && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &> 1 \\ \\ && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} & > \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = 1 \end{align*} So we're done.

Solution:

1. \(\,\) \begin{align*} && x &= \frac1{1-u} \\ \Rightarrow && \d x &= \frac{1}{(1-u)^2} \d u \\ && I &= \int \frac{1}{x^{\frac32}(x-1)^{\frac12} } \d x \\ &&&= \int \frac1{(1-u)^{-\frac32}u^{\frac12}(1-u)^{-\frac12}} (1-u)^{-2} \d u \\ &&&= \int u^{-\frac12} \d u \\ &&&= 2\sqrt{u} + C \\ &&&= 2\sqrt{1-\frac{1}{x}} + C \end{align*}
2. \(\,\) \begin{align*} && J &= \int \frac{1}{(x-2)^{\frac32}(x+1)^{\frac12}} \d x \\ y = x+1: &&&= \int \frac{1}{(y-3)^{\frac32}y^{\frac12}} \d y \\ y = 9(3-u)^{-1}: &&&= \int \frac1{\left (9(3-u)^{-1}-3 \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \int \frac1{\left (3u(3-u)^{-1} \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \frac{1}{\sqrt3} \int u^{-\frac32} \d u \\ &&&= -\frac2{\sqrt3} u^{-\frac12} + C \\ &&&= -\frac2{\sqrt3} \sqrt{\frac{y}{3(y-3)}} + C \\ &&&= -\frac2{3} \sqrt{\frac{x+1}{x-2}} + C \\ \end{align*}
3. \(\,\) \begin{align*} && K &= \int_2^{\infty} \frac{1}{(x-1)(x-2)^{\frac12}(3x-2)^{\frac12}} \d x \\ y = x - 1: &&&=\int_{y=1}^{\infty} \frac{1}{y(y-1)^{\frac12}(3y+1)^{\frac12}} \d y \\ y = (1-u)^{-1}: &&&= \int_{u=0}^{u=1} \frac{1}{(1-u)^{-1}(u(1-u)^{-1})^{\frac12}((4-u)(1-u)^{-1})^{\frac12}} \frac{1}{(1-u)^2} \d u \\ &&&= \int_0^1 \frac{1}{u^{\frac12}(4-u)^{\frac12}} \d u \\ &&&= \int_0^1 \frac{1}{\sqrt{4-(u-2)^2}} \d u \\ &&&= \left [-\sin^{-1} \left ( \frac{2-u}{2} \right) \right]_0^1 \\ &&&= \sin^{-1} 1 - \sin^{-1} \tfrac12 \\ &&&= \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \end{align*}

Solution: \begin{align*} \int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{1 - \sin^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{\cos^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \sec^2 x - \sec x \tan x dx \\ &= \left [\tan x-\sec x \right]_0^{\frac{1}{4}\pi} \\ &= 2 - \frac{1}{\sqrt{2}} \end{align*} \begin{align*} \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} \d x &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1-\sec x}{1 - \sec^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{\sec x-1}{\tan^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \cot x \cosec x-\cot^2 x\d x \\ &= \left [ -\cosec x +x+\cot x\right]_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \\ &= \l -\frac{2}{\sqrt3}+\frac{\pi}{3}+\frac{1}{\sqrt{3}}\r - \l-\sqrt{2}+\frac{\pi}{4}+1 \r \\ &= \frac{\pi}{12}-\frac{1}{\sqrt{3}}+\sqrt{2}-1 \end{align*} \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{(1-\sin^2 x)^2} \d x \\ &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{\cos^4 x} \d x \\ \end{align*} Splitting this up into: \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{-2\sin x}{\cos^4 x} \d x &= -\frac23 \left [ \frac{1}{\cos^3 x}\right]_0^{\frac{1}{3}\pi} \\ &= -\frac{16}3+\frac23 \\ &= -\frac{14}3 \end{align*} and \begin{align*} && \int_0^{\frac{1}{3}\pi} \frac{1+\sin^2x}{\cos^4 x} \d x &= \int_0^{\frac{1}{3}\pi} (\sec^2 x + \tan^2 x) \sec^2 x \d x \\ &&&= \int_0^{\frac{1}{3}\pi} (1+ 2\tan^2 x) \sec^2 x \d x \\ u = \tan x, \d u = \sec^2 x \d x&&&= \int_0^{\sqrt{3}}(1+2u^2) \d u \\ &&&= \left [u + \frac23 u^3 \right]_0^{\sqrt{3}} \\ &&&= \sqrt{3} + 2\sqrt{3} \\ &&&= 3\sqrt{3} \end{align*} And so our complete integral is: \[ \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x = 3\sqrt{3} - \frac{14}3\]

Solution:

1. Let \(2t = u\), \begin{align*} \int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\ &= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\ &= 2f\l\frac{x}{2}\r \end{align*}
2. Let \(v = u-2\), \begin{align*} \int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\ &= 2 f \l \frac{x+2}{2} \r \end{align*}
3. Let \(v = u-2, \d v = \d u\) \begin{align*} \int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\ &= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\ &= 3 f \l \frac{x-2}{3} \r \end{align*}
4. Let \(v = u^2, \d v = 2u \d u\)\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\ &= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\ &= f \l \frac{4+2}{2} \r - f \l \frac{3}{2} \r \\ &= f(3) - f(\frac32) \end{align*}

Solution:

1. \(\,\)
   

   + - ↺
2. \(\,\) \begin{align*} && y &= \frac{cx}{\sqrt{x^2+p}} \\ && \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\ &&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\ && I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\ &&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p} }} \d y \\ &&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\ &&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\ p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x \end{align*} When \(b = \sqrt{3}, c = \sqrt{2}\) \begin{align*} && I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\ &&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\ &&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\ &&&= \frac{\pi}{3\sqrt{3}} \end{align*} \begin{align*} && I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\ &&&= \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\ &&&= I_1 = \frac{\pi}{3\sqrt{3}} \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\ u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2} \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\ v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\ &&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\ &&&= \frac{\pi}{4} \end{align*}

Solution:

1. Let \(x = \frac{pz+q}{z+1}\) then \begin{align*} && 0 &= x^3-3pqx+pq(p+q) \\ &&&= \left ( \frac{pz+q}{z+1} \right)^3 - 3pq \left ( \frac{pz+q}{z+1} \right) + pq(p+q) \\ &&&= \frac{(pz+q)^3-3pq(pz+q)(z+1)^2+pq(p+q)(z+1)^3}{(z+1)^3} \\ &&&= \frac{1}{(z+1)^3} \Big ((p^3+pq(p+q)-3p^2q)z^3 + (3p^2q-6p^2q+3pq^2+3p^2q+3pq^2)z^2 + \\ &&&\qquad \qquad\quad\quad +(3pq^2-3p^2q-6pq^2+3p^2q+3qp^2)z+(q^3-3pq^2+p^2q+pq^2) \Big ) \\ &&&= \frac{(p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2)}{(z+1)^3} \\ \Rightarrow && 0 &= (p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2) \\ &&&= p(p-q)^2z^3 + q(p-q)^2 \\ \Rightarrow && 0 &= pz^3 + q \end{align*}
2. We would like to find \(pq = c\) and \(pq(p+q) = d\), so \(p\) and \(q\) are roots of the quadratic \(x^2-\frac{d}{c}x + c = 0\), which has distinct real roots if \(\Delta = \frac{d^2}{c^2}-4c > 0 \Rightarrow d^2>4c^3\)
3. Note that \(c = -2, d = -2\) so \begin{align*} && 0 &= x^3+6x-2 \\ \text{consider} && 0 &= X^2-X-2 \\ && &= (X+1)(X-2) \\ \Rightarrow && p = -1, &q = 2\\ \Rightarrow && 0 &= x^3-3\cdot 2 \cdot(-1) x + 2\cdot(-1) \cdot(-2+1) \\ \Rightarrow && 0 &= -z^3+2 \\ \Rightarrow && z &= \sqrt[3]{2} \\ \Rightarrow && \frac{-z+2}{z+1} &= \sqrt[3]{2} \\ \Rightarrow && -z+2 &= \sqrt[3]{2} z + \sqrt[3]{2} \\ \Rightarrow && z &= \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1} \end{align*}
4. \(\,\) \begin{align*} && 0 &= x^3 - 3p^2x + 2p^3 \\ &&&= (x-p)(x^2+px-2p^2) \\ &&&=(x-p)^2(x+2p)\\ \Rightarrow && x &= p, p, -2p \end{align*} Therefore if we have a repeated root to our associated quadratic we can find a cubic of the form \(x^3-3p^2x+2p^3\), but we know this has roots we can find.

Solution: \begin{align*} x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\ &= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\ &= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\ &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \end{align*}

1. If \(x_{n+2} = x_n\) then \begin{align*} && x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\ \Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\ &&&= (a+b)(x_n^2+(a-b)x_n + 1) \end{align*} If \(x_{n+1} = x_n\) then \(x_n^2+(a-b)x_n + 1\) and since our sequence has period \(2\) rather than \(1\) it must be the case this is non-zero. Therefore \(a+b =0\).
2. \begin{align*} x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\ &= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\ &= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \end{align*} If \(x_{n+4} = x_n\) then \begin{align*} x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\ 0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1) \end{align*} Since we do not want \(x_n\) to be periodic with period \(1\) we must have the quadratic in \(x_n\) \(\neq 0\). If \(a+b = 0\) then \(x_n\) is periodic with period \(2\) since \(x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n\). Therefore it is necessary that \(a^2+b^2-2 = 0\). If \(a^2+b^2-2= 0\) then \begin{align*} x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\ &= x_n \end{align*} Therefore it is sufficient too. So our conditions are \(a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0\) and \(a^2+b^2-2 = 0\)

Solution:

1. Claim: \(S_n \leq 2\sqrt{n} -1\). Proof: (By induction) (Base case: \(n = 1\)). \(\frac{1}{\sqrt{1}} \leq 1 = 2 \cdot \sqrt1 - 1\). Therefore the base case is true. (Inductive step): Suppose our result is true for \(n = k\). Then consider \(n = k+1\). \begin{align*} && \sum_{r=1}^{k+1} \frac{1}{\sqrt{r}} &=\sum_{r=1}^{k} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{k+1}} \\ &&&\leq 2\sqrt{k} - 1 + \frac{1}{\sqrt{k+1}} \\ &&&= \frac{2 \sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}} - 1 \\ &&&\underbrace{\leq}_{AM-GM} \frac{(k+k+1)+1}{\sqrt{k+1}} - 1 \\ &&&=\frac{2(k+1)}{\sqrt{k+1}} - 1 \\ &&&= 2\sqrt{k+1}-1 \end{align*} Therefore, since if our statement is true for \(n = k\), it is also true for \(n = k+1\). By the principle of mathematical induction we can say that it is true for all \(n \geq 1, n \in \mathbb{Z}\)
2. Claim: \((4k+1)\sqrt{k+1} > (4k+3)\sqrt k\,\) for \(k\ge0\,\) Proof: \begin{align*} && (4k+1)\sqrt{k+1} &> (4k+3)\sqrt k \\ \Leftrightarrow && (4k+1)^2(k+1) &> (4k+3)^2k \\ \Leftrightarrow && (16k^2+8k+1)(k+1) &> (16k^2 + 24k+9)k \\ \Leftrightarrow && 16 k^3 + 24 k^2 + 9 k +1&> 16k^3 + 24k^2+9k \end{align*} But this last inequality is clearly true, hence our original inequality is true. Suppose \(S_n \geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C\), then adding \(\frac{1}{\sqrt{n+1}}\) to both sides we have: \begin{align*} S_{n+1} &\geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C + \frac{1}{\sqrt{n+1}} \\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} +2(\sqrt{n} - \sqrt{n+1})\\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} -\frac{2}{\sqrt{n+1} + \sqrt{n}}\\ \end{align*} Therefore as long as the inequality is satisfied for \(n=1\), ie \(1 \geq 2\sqrt{1} + \frac{1}{2 \sqrt{1}} - C = \frac52 - C \Rightarrow C \geq \frac32\)

Solution:

1. \(\,\) \begin{align*} && T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\ &&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\ u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\ &&&= T_\infty - T(x^{-1}) \end{align*}
2. Let \(v = \frac{u+a}{1-au}\) then \begin{align*} && \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\ &&&= \frac{1-au+au+a^2}{(1-au)^2} \\ &&&= \frac{1+a^2}{(1-au)^2} \\ \\ && \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\ &&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+a^2}{(1-ua)^2} \end{align*} if \(a > 0, x < \frac1a\) then \begin{align*} && T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\ &&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\ \\ \Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\ &&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1}) \end{align*} \(b > 0, y > \frac1b\) then \(y> 0, b > \frac1y\) (same as letting \(x = \frac1y, a = \frac1b\) \begin{align*} && T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\ \Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\ \end{align*}
3. Letting \(y = b = \sqrt{3}\) in the final equation \begin{align*} && T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\ &&&= 2T_\infty - 2T(\sqrt{3}) \\ \Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty \end{align*} Let \(x = \sqrt2 - 1, a = 1\) so, \begin{align*} && T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\ &&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\ &&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\ &&&= T(\sqrt{2}+1) - T(1) \\ &&&= T_\infty - T(\sqrt2-1)-T(1) \\ \Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\ && T(1) &= T_\infty - T(1) \\ \Rightarrow && T(1) &= \frac12 T_\infty \\ \Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\ &&&= \frac14 T_\infty \end{align*}

Solution:

1. \(n=1\): \begin{align*} && p_1(x) &= (x+1)^2 - 3x(x^2+x+1)^0 \\ &&&= x^2+2x+1-3x \\ &&&= x^2-x+1\\ && q_1(x) &= \frac{x^3+1}{x+1} \\ &&&= x^2-x+1 = p_1(x) \\ \\ && p_2(x) &= (x+1)^4-5x(x^2+x+1)^1 \\ &&&= x^4+4x^3+6x^2+4x+1 - 5x^3-5x^2-5x \\ &&&= x^4-x^3+x^2-x+1 \\ &&q_2(x) &= \frac{x^5+1}{x+1} \\ &&&= x^4-x^3+x^2-x+1 = p_2(x) \\ \\ && p_3(x) &= (x+1)^6-7x(x^2+x+1)^2 \\ &&&= x^6+6x^5+15x^4+20x^3+15x^2+6x+1 - 7x(x^4+2x^3+3x^2+2x+1) \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 \\ && q_3(x) &= \frac{x^7+1}{x+1} \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 = p_3(x) \\ \\ && p_4(1) &= 2^8 - 9 \cdot 1 \cdot 3^3 \\ &&&= 256 - 243 = 13 \\ && q_4(1) &= \frac{2}{2} = 1 \neq 13 \end{align*}
2. • \(\bf (a)\) \(\,\) \begin{align*} && \frac{300^3+1}{300+1} &= (300+1)^2 - 3 \cdot 300 \\ &&&= 301^2 - 30^2 \\ &&&= 271 \cdot 331 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && \dfrac {7^{49}+1}{7^7+1} &= (7^7+1)^6 - 7 \cdot 7^7 \cdot (7^2+7+1)^2 \\ &&&= (7^7+1)^6 - 7^8 \cdot (7^2+7+1)^2 \\ &&&= ((7^7+1)^3 - 7^4(7^2+7+1)) \cdot ((7^7+1)^3 + 7^4(7^2+7+1)) \end{align*}